MCQ
$\lim_{x \rightarrow \frac{\pi}{8}} \frac{sinx - sin\frac{\pi}{8}}{8x-\pi}=$ .............
- A$\frac{1}{16}(2+\sqrt{2})$
- B$\frac{1}{16}(\sqrt{2-\sqrt{2}} )$
- C$\frac{1}{16}(2-\sqrt{2})$
- ✓$\frac{1}{16}\left(\sqrt{2+\sqrt{2}}\right)$
$\lim_{x \rightarrow \frac{\pi}{8}} \frac{\sin x-\sin \frac{\pi}{8}}{8x-\pi}$
$\left(\frac{0}{0}\right)from.$
$\lim_{x \rightarrow \frac{\pi}{8}} \frac{\cos x}{8}$
$=\frac{1}{16}\left(\sqrt{2+\sqrt{2}}\right)$
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